Solve the system of equations. $\begin{aligned} &15x+31y = -3 \\\\ &x=-y+3 \end{aligned}$ $x=$
Solution: We are given that ${x}={-y+3}$. Let's substitute this expression into the first equation and solve for $y$ as follows: $\begin{aligned} 15{x}+31y &= -3\\\\ 15\cdot({-y+3})+31y&=-3\\\\ -15y+45+31y&=-3\\\\ 16y&=-48\\\\ y&=-3 \end{aligned}$ Since we now know that ${y}={-3}$, we can substitute this value in the second equation to solve for $x$ as follows: $\begin{aligned} x &= -{y} +3\\\\ x&=-({-3})+3\\\\ x&=6 \end{aligned}$ This is the solution of the system: $\begin{aligned} &x = 6 \\\\ &y=-3 \end{aligned}$